Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:


import java.util.Scanner;

public class Main {
    public static double[][] probs;

    public static void main(String[] args){
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        probs = new double[n+1][n+1];
        for(int i = 0; i <= n; i++){
            for(int j = 0; j <= n; j++){
                probs[i][j] = 0;
            }
        }
        probs[0][0] = 1;
        double[] coins = new double[n+1];
        for(int i = 1; i <= n; i++){
            coins[i] = scanner.nextDouble();
        }
        coins(coins, n);
        double res = 0;
        for(int i = n/2 +1; i <= n; i++){
            res += probs[i][n];
        }
        System.out.println(res);
    }

    public static void coins(double[] coins, int n){
        for(int c = 1; c <= n; c++){
            for(int h = 0; h <= c; h++){
                if(h==c){
                    probs[h][c] = probs[h-1][c-1] * coins[c];
                }else if(h==0){
                    probs[h][c] = probs[h][c-1] * (1-coins[c]);
                }else{
                    probs[h][c] = probs[h][c-1] * (1-coins[c]) + probs[h-1][c-1] * coins[c];
                }
            }
        }
    }
}